Suppose you carry out a significance test of H0: μ = 3.5 versus Ha: μ < 3.5 based on sample size n = 17 and obtain t = - 3.4. Find the p-value for this test. What conclusion can you draw at the 5% significance level? Explain. a. The p-value is 0.4982. We reject H0 at the 5% significance level because the p-value 0.4982 is greater than 0.05. b. The p-value is 0.4982. We fail to reject H0 at the 5% significance level because the p-value 0.4982 is greater than 0.05. c. The p-value is 0.5018. We fail to reject H0 at the 5% significance level because the p-value 0.5018 is greater than 0.05. d. The p-value is 0.0018. We reject H0 at the 5% significance level because the p-value 0.0018 is less than 0.05. e. The p-value is 0.0018. We fail to reject H0 at the 5% significance level because the p-value 0.0018 is less than 0.05.

Question

Mathematics

Malcolm

12 January, 06:10

Suppose you carry out a significance test of H0: μ = 3.5 versus Ha: μ < 3.5 based on sample size n = 17 and obtain t = - 3.4. Find the p-value for this test. What conclusion can you draw at the 5% significance level? Explain. a. The p-value is 0.4982. We reject H0 at the 5% significance level because the p-value 0.4982 is greater than 0.05. b. The p-value is 0.4982. We fail to reject H0 at the 5% significance level because the p-value 0.4982 is greater than 0.05. c. The p-value is 0.5018. We fail to reject H0 at the 5% significance level because the p-value 0.5018 is greater than 0.05. d. The p-value is 0.0018. We reject H0 at the 5% significance level because the p-value 0.0018 is less than 0.05. e. The p-value is 0.0018. We fail to reject H0 at the 5% significance level because the p-value 0.0018 is less than 0.05.

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Ashtyn Benitez · Answer 1

We reject H0 at the 5% significance level because the p-value 0.0018 is less than 0.05 is the correct answer here.

Step-by-step explanation:

For n - 1 = 16 degrees of freedom, we get from the t distribution tables for this one tailed test the p-value as:

p = P (t16 < - 3.4) = 0.0018

As the p-value here is 0.0018 < 0.05 which is the level of significance, therefore the test is significant and we can reject the null hypothesis here. Therefore The p-value is 0.0018. We reject H0 at the 5% significance level because the p-value 0.0018 is less than 0.05 is the correct answer here.